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Let $G$ is a $p-$group and $H$ a proper subgroup of $G$, $|H|=p^s$. Show that $G$ has a subgroup $K$ such that $H \subset K$ and $|K|=p^{s+1}$.

I try to find a subgroup $K$ of $G$, $|K|=p^t$ with some $t>s$. Let $K=\langle g,H\rangle$ with some $g\in G,g\not\in H$, we have ${|g|=p^n,|H|=p^s}\Rightarrow|K|=p^n\times p^s$. Is this true? Can you help me!

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Auch! You must really enhance your accept rate, otherwise many people won't feel like trying to help you out... –  DonAntonio Jan 1 '13 at 14:48
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For this we can use that a $p$-group has a non-trivial center and proceed by induction on the order. If the subgroup does not contain the center then its normalizer does and the normalizer is therefore a strictly larger subgroup. If it does contain the center then you can mod out by a central subgroup of order $p$ and use the induction.

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@ Tobias: Ok, a finite $p-$group $G$ has a non-trivial center. But is it true in case G is infinite? –  tlquyen Jan 2 '13 at 14:28
    
The result you want to show does not hold for arbitrary $p$-groups. For example, there exists an infinite group where all proper subgroups have order $p$ for a fixed prime $p$. –  Tobias Kildetoft Jan 2 '13 at 15:16
    
Oh, this is an exercise in my final test. Can you give me some details about the example that you say about. Thank you! –  tlquyen Jan 2 '13 at 15:44
    
The infinite groups I mentioned are called the Tarski monster groups (there is a short article about them on Wikipedia). –  Tobias Kildetoft Jan 2 '13 at 15:47
    
Tobias, thank you very much. –  tlquyen Jan 2 '13 at 16:02
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